I'm in the process of building a dual purpose power supply. The supply
will have two outputs. One will be 24VAC rated at 4.0 A. The load on
this supply will probably never exceed 2.0 A. The second supply will
consist of a bridge rectifier off the 24V tap connected to a suitable
filter cap of say 1000UF. This filtered DC will then be connected to a
small surplus 12V regulator board and heat sink assembly which by the
looks of it can handle 5A or better. I plan on mounting this regulator
in the cabinet and providing a terminal strip for the 12V output. The
DC load will probably never exceed .500 - .750 A.

Now here is the dilemma. I haven't tried to put all this together yet,
but I know that once I rectify and filter the 24VAC I'll probably end
up with something like 30VDC out. The regulator uses a ua723, a TO3
and a smaller TO transistor as well a whole bunch of discrete
components. I have no specs on this regulator, but it is well built,
appears to be commercial grade, and although it might handle it fine
I'd feel a lot better hitting it with something like 18VDC instead of
30. I would hate to blow it up trying to see if it would work on the
higher input.

I could add some series resistance either before or after the bridge,
but the voltage drop across this resistance would vary depending on
the total load and I'm not sure how well such a scheme would work. So
I was thinking about employing a voltage divider at the output of the
filter. The resistance ratio would be easy to figure out, however I'm
just not sure how to determine the optimum resistor values. Does this
seem like a viable plan, or perhaps someone my have other thoughts as
to how to address this? If someone could please help me with this I
would be very grateful. Thanks, Lenny

On 2/18/2013 9:49 AM, klem kedidelhopper wrote:
I'm in the process of building a dual purpose power supply. The supply
will have two outputs. One will be 24VAC rated at 4.0 A. The load on
this supply will probably never exceed 2.0 A. The second supply will
consist of a bridge rectifier off the 24V tap connected to a suitable
filter cap of say 1000UF. This filtered DC will then be connected to a
small surplus 12V regulator board and heat sink assembly which by the
looks of it can handle 5A or better. I plan on mounting this regulator
in the cabinet and providing a terminal strip for the 12V output. The
DC load will probably never exceed .500 - .750 A.

Now here is the dilemma. I haven't tried to put all this together yet,
but I know that once I rectify and filter the 24VAC I'll probably end
up with something like 30VDC out. The regulator uses a ua723, a TO3
and a smaller TO transistor as well a whole bunch of discrete
components. I have no specs on this regulator, but it is well built,
appears to be commercial grade, and although it might handle it fine
I'd feel a lot better hitting it with something like 18VDC instead of
30. I would hate to blow it up trying to see if it would work on the
higher input.

I could add some series resistance either before or after the bridge,
but the voltage drop across this resistance would vary depending on
the total load and I'm not sure how well such a scheme would work. So
I was thinking about employing a voltage divider at the output of the
filter. The resistance ratio would be easy to figure out, however I'm
just not sure how to determine the optimum resistor values. Does this
seem like a viable plan, or perhaps someone my have other thoughts as
to how to address this? If someone could please help me with this I
would be very grateful. Thanks, Lenny

Not clear exactly what you're doing, but the devil is in the details.
Depending on the line voltage and the design of the transformer, but
the unloaded voltage of the transformer may result significantly higher
than 30V.

The word "24V tap" begs the question of how the transformer
is configured to get the 30VDC.

You really need to characterize the 12V board. It doesn't take a
"whole bunch" of discretes to make a 723 supply. There may
be other things going on. You don't know
what input min/max it can tolerate. You're gonna have a wide
range in input voltage depending on line voltage and load current.
And a couple of volts of ripple. Once you figure out how much
headroom you have, insert a zener in the power input.
You can make one from a small zener and a power transistor.
Doesn't have to be all that accurate.

And there are a lot of potential gotchas using an unknown
board. For example, I build supplies that sense output
current in the negative lead. They run from floating transformer
windings. If I were to try to use the AC for some other purpose,
stuffing current into my floating winding relative to ground
would make a real mess of things.

Making a dual-purpose power supply where both purposes are known
is much easier than building an dual-output supply where
none of the "purposes" are known in advance.

On Mon, 18 Feb 2013 09:49:18 -0800 (PST), klem kedidelhopper
wrote:

I'm in the process of building a dual purpose power supply. The supply
will have two outputs. One will be 24VAC rated at 4.0 A. The load on
this supply will probably never exceed 2.0 A. The second supply will
consist of a bridge rectifier off the 24V tap connected to a suitable
filter cap of say 1000UF. This filtered DC will then be connected to a
small surplus 12V regulator board and heat sink assembly which by the
looks of it can handle 5A or better. I plan on mounting this regulator
in the cabinet and providing a terminal strip for the 12V output. The
DC load will probably never exceed .500 - .750 A.

Now here is the dilemma. I haven't tried to put all this together yet,
but I know that once I rectify and filter the 24VAC I'll probably end
up with something like 30VDC out. The regulator uses a ua723, a TO3
and a smaller TO transistor as well a whole bunch of discrete
components. I have no specs on this regulator, but it is well built,
appears to be commercial grade, and although it might handle it fine
I'd feel a lot better hitting it with something like 18VDC instead of
30. I would hate to blow it up trying to see if it would work on the
higher input.

I could add some series resistance either before or after the bridge,
but the voltage drop across this resistance would vary depending on
the total load and I'm not sure how well such a scheme would work. So
I was thinking about employing a voltage divider at the output of the
filter. The resistance ratio would be easy to figure out, however I'm
just not sure how to determine the optimum resistor values. Does this
seem like a viable plan, or perhaps someone my have other thoughts as
to how to address this? If someone could please help me with this I
would be very grateful. Thanks, Lenny
The best aproach would be to use a transformer with both 12V and 24V
windings. Feeding 24VAC into a bridge rectifier will give you about
32 - 36 volts, depending on load. Using a rough approximation, your
regulator would be dissipating 15 watts. I'd also boost the value of
the main filter cap by at least a value of 10.

An alternative approach may be to use a 7812 series regulator as an
adjustable regulator. By using a voltage divider between the output,
the reference (ground) pin, and true ground you could have an
intermediate 18V regulated output to feed to your regulator. Check
the 7812 data sheets for voltage readings and how to do this.

If you have a variable power supply, or a variable transformer it may
be possible to evaluate your regulator. Use a 100 ohm resistor
between the power supply and the input pin of the regulator. Measure
the voltage across the resistor while slowly increasing the voltage
into the regulator. If a slight increase in input voltage results in
a signiificant increase in voltage across the 100 ohm resistor you are
above the maximum input voltage for the regulator.

PlainBill

klem kedidelhopper wrote:

I'm in the process of building a dual purpose power supply. The supply
will have two outputs. One will be 24VAC rated at 4.0 A. The load on
this supply will probably never exceed 2.0 A. The second supply will
consist of a bridge rectifier off the 24V tap connected to a suitable
filter cap of say 1000UF. This filtered DC will then be connected to a
small surplus 12V regulator board and heat sink assembly which by the
looks of it can handle 5A or better. I plan on mounting this regulator
in the cabinet and providing a terminal strip for the 12V output. The
DC load will probably never exceed .500 - .750 A.

Now here is the dilemma. I haven't tried to put all this together yet,
but I know that once I rectify and filter the 24VAC I'll probably end
up with something like 30VDC out. The regulator uses a ua723, a TO3
and a smaller TO transistor as well a whole bunch of discrete
components. I have no specs on this regulator, but it is well built,
appears to be commercial grade, and although it might handle it fine
I'd feel a lot better hitting it with something like 18VDC instead of
30. I would hate to blow it up trying to see if it would work on the
higher input.

I could add some series resistance either before or after the bridge,
but the voltage drop across this resistance would vary depending on
the total load and I'm not sure how well such a scheme would work. So
I was thinking about employing a voltage divider at the output of the
filter. The resistance ratio would be easy to figure out, however I'm
just not sure how to determine the optimum resistor values. Does this
seem like a viable plan, or perhaps someone my have other thoughts as
to how to address this? If someone could please help me with this I
would be very grateful. Thanks, Lenny

It can handle up to 40volts on the input..

http://www.ti.com/lit/ds/symlink/lm723.pdf

There is PDF with examples at the bottom for its use..

Jamie

klem kedidelhopper wrote:
The second supply will consist of a bridge rectifier off the 24V tap
connected to a suitable filter cap of say 1000UF.

With a full-wave bridge rectifier, if you use an 8300 uF capacitor, the
volts of ripple will equal the amps of load current. (Don Lancaster
taught me this.) The proposed 1000 uF capacitor will have about 8 volts
of ripple at a 1 amp load current. This is still above what the
regulator probably needs as a minimum input, but you might want to
consider a bigger capacitor.

I could add some series resistance either before or after the bridge,
but the voltage drop across this resistance would vary depending on
the total load and I'm not sure how well such a scheme would work.

I have a USB power supply that plugs in to the 12 V in my car that does
this. The 12 V DC goes through a power resistor (around 5 W rating) and
then into the input of a 7805 regulator in a TO-220 case. I'm not sure
if they did this to move some of the heat dissipation from the regulator
to the resistor, or just wanted to provide some current limiting for
some failure modes of the regulator, or what. I've never put a meter on
it but it seems to work OK. (The USB socket provides 5 V DC at 0.5 A,
or 2.5 W.)

If you don't have any better specs on what the parts will do, I'd try it
and see. Install the transformer, bridge rectifier, and capacitor, but
leave out the regulator. Load the 24 VAC output to 4 A. (If you don't
have some power resistors sitting around, car tail light bulbs in series
are good for this and you probably already own some.) Load the output
of the bridge rectifier to 1 A. Plug the transformer in to the lowest
line voltage you expect it to work at (use a Variac if you have one).
Measure the DC at the output of the rectifier - that is the lowest DC
input to the regulator you can expect. If you want, you can then
disconnect the load on the 24 VAC output and on the rectifier, plug into
the highest line voltage you expect (Variac again), and measure the DC
again, to get the highest DC input to the regulator that you can expect.

Then, you need to know the minimum DC input you can have to your
regulator board for it to still regulate. If you already have a
variable DC power supply, this is easy to figure out. If you don't, you
have to guess; if it was a 7812 the standard answer is that you need
about 13.5 to 14 V DC minimum for 12 V DC out. Your board may be
designed for more than this, though.

Once you know the lowest DC voltage you can expect from the rectifier,
and the lowest DC voltage you can supply to the regulator, you can
figure out how many volts maximum you have to drop in the resistor.

Just on paper, if I guesstimate a minimum DC input to the regulator of
14 V, and a 1 A load, I have...

Peak secondary voltage 24 V * 1.414 = 33.9 V
Minus two diode drops 33.9 V - 2 V = 31.9 V
Ripple with 1000 uF @ 1 A 31.9 V - 8 V = 23.9 V
Minus regulator minimum 23.9 V - 14 V = 9.9 V
Resistor needed 9.9 V / 1 A = 9.9 ohms
9.9 V * 1 A = 9.9 watts

I would probably then look for something like an 8 ohm, 15 watt or
better resistor. In perfect conditions, this would provide a little
more DC to the input of the regulator, and in the real world, it would
also account for getting less than 24 V DC from the transformer,
slightly more diode drop, slightly worse ripple, etc.

So I was thinking about employing a voltage divider at the output of
the filter.

The series resistor *is* a voltage divider. The regulator is the
bottom resistor in the divider.

Matt Roberds

"klem kedidelhopper" wrote in message
...
I'm in the process of building a dual purpose power supply. The supply
will have two outputs. One will be 24VAC rated at 4.0 A. The load on
this supply will probably never exceed 2.0 A. The second supply will
consist of a bridge rectifier off the 24V tap connected to a suitable
filter cap of say 1000UF. This filtered DC will then be connected to a
small surplus 12V regulator board and heat sink assembly which by the
looks of it can handle 5A or better. I plan on mounting this regulator
in the cabinet and providing a terminal strip for the 12V output. The
DC load will probably never exceed .500 - .750 A.

Now here is the dilemma. I haven't tried to put all this together yet,
but I know that once I rectify and filter the 24VAC I'll probably end
up with something like 30VDC out. The regulator uses a ua723, a TO3
and a smaller TO transistor as well a whole bunch of discrete
components. I have no specs on this regulator, but it is well built,
appears to be commercial grade, and although it might handle it fine
I'd feel a lot better hitting it with something like 18VDC instead of
30. I would hate to blow it up trying to see if it would work on the
higher input.

I could add some series resistance either before or after the bridge,
but the voltage drop across this resistance would vary depending on
the total load and I'm not sure how well such a scheme would work. So
I was thinking about employing a voltage divider at the output of the
filter. The resistance ratio would be easy to figure out, however I'm
just not sure how to determine the optimum resistor values. Does this
seem like a viable plan, or perhaps someone my have other thoughts as
to how to address this? If someone could please help me with this I
would be very grateful. Thanks, Lenny

There's a good chance that there will be an electrolytic decoupling cap hung
across the supply input of your regulator board. Might be anything from 10uF
to 220uF. If you can find such a cap, look at it's voltage rating. If it's
say 25v then the chances are that the board is not expecting more than 20v
of input. If it's a 35v part, then probably 30v tops.

You can drop a constant amount in a simple fashion by using a series zener
rather than a series resistor, which would, as you surmise, produce a
variable drop depending on load. You just put the zener in the line pointing
'backwards'. However, bear in mind that the power rating of the zener has to
be sufficient, so if you were looking to drop say 10v at 0.5 A, you would be
looking at a 5 watt minimum zener. As an alternative, 78xx regulators come
in many voltages, including some quite high ones (I have 18s and 24s). Most
will do at least an amp, and some versions more. Bear in mind if going down
this route that you will need to heatsink the device, and you will need to
add decoupling caps close to the pins to avoid instability problems.

Arfa

"klem kedidelhopper"

Now here is the dilemma. I haven't tried to put all this together yet,
but I know that once I rectify and filter the 24VAC I'll probably end
up with something like 30VDC out.

** Likely about 35VDC unloaded


I could add some series resistance either before or after the bridge,
but the voltage drop across this resistance would vary depending on
the total load and I'm not sure how well such a scheme would work.


** A string of say 20 x 3A diodes is the cheapest solution.

Wired in series with the "+" terminal of the bridge, all mounted on a tag
strip.

Drops the DC level by about 12 to 15 volts, depending on the load.


..... Phil

With a full-wave bridge rectifier, if you use an 8300 uF capacitor, the
volts of ripple will equal the amps of load current. (Don Lancaster
taught me this.)


** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a full
wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1 amp.



..... Phil

On Feb 19, 12:51*am, "Phil Allison" wrote:




With a full-wave bridge rectifier, if you use an 8300 uF capacitor, the
volts of ripple will equal the amps of load current. *(Don Lancaster
taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a full
wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1 amp.

.... *Phil

This is some great advice. Thank you everyone for all your input. I
especially like the diode and Zener ideas. They/re cheap and should
work well. I've got lots of diodes around here and I'm going to
experiment with them. However after all this I remembered that the
transformer is center tapped! At the time, I simply wire nutted the
tap, tucked it down under the transformer and forgot about it. How
this simple fact and the possibilities it presents slipped my mind is
anyone's guess. I've been building this thing in the evenings after my
usual work and perhaps I've been tired. The transformer measures 25.2V
at 120V line unloaded. Loading the transformer in this fashion will
probably unbalance it somewhat and drop the 24V a bit as well but I
don't see it as a real problem. I should be able to use one side of
the secondary and the tap now get at least 15 -17 VDC out of the
bridge and filter, and that will provide a healthier input to my
regulator. Lenny

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:
On Feb 19, 12:51 am, "Phil Allison" wrote:




With a full-wave bridge rectifier, if you use an 8300 uF capacitor, the
volts of ripple will equal the amps of load current. (Don Lancaster
taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a full
wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1 amp.

.... Phil

This is some great advice. Thank you everyone for all your input. I
especially like the diode and Zener ideas. They/re cheap and should
work well. I've got lots of diodes around here and I'm going to
experiment with them. However after all this I remembered that the
transformer is center tapped! At the time, I simply wire nutted the
tap, tucked it down under the transformer and forgot about it. How
this simple fact and the possibilities it presents slipped my mind is
anyone's guess. I've been building this thing in the evenings after my
usual work and perhaps I've been tired. The transformer measures 25.2V
at 120V line unloaded. Loading the transformer in this fashion will
probably unbalance it somewhat and drop the 24V a bit as well but I
don't see it as a real problem. I should be able to use one side of
the secondary and the tap now get at least 15 -17 VDC out of the
bridge and filter, and that will provide a healthier input to my
regulator. Lenny

No need to unbalance the transformer, just use 2 diodes and use the
center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

Mikek

On Feb 20, 6:06*pm, amdx wrote:
On 2/19/2013 5:01 PM, klem kedidelhopper wrote:







On Feb 19, 12:51 am, "Phil Allison" wrote:


With a full-wave bridge rectifier, if you use an 8300 uF capacitor, the
volts of ripple will equal the amps of load current. *(Don Lancaster
taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a full
wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1 amp.

.... *Phil

This is some great advice. Thank you everyone for all your input. I
especially like the diode and Zener ideas. They/re cheap and should
work well. I've got lots of diodes around here and I'm going to
experiment with them. However after all this I remembered that the
transformer is center tapped! *At the time, I simply wire nutted the
tap, tucked it down under the transformer and forgot about it. How
this simple fact and the possibilities it presents slipped my mind is
anyone's guess. I've been building this thing in the evenings after my
usual work and perhaps I've been tired. The transformer measures 25.2V
at 120V line unloaded. Loading the transformer in this fashion will
probably unbalance it somewhat and drop the 24V a bit as well but I
don't see it as a real problem. I should be able to use one side of
the secondary and the tap now get at least 15 -17 VDC out of the
bridge and filter, and that will provide a healthier input to my
regulator. Lenny

* *No need to unbalance the transformer, just use 2 diodes and use the
center tap as negative. See Here,

*http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

* * * * * * * * * * * * * * * * Mikek

But won't that still give me 36VDC into my regulator? Lenny

"klem kedidelhopper"

No need to unbalance the transformer, just use 2 diodes and use the
center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

Mikek

But won't that still give me 36VDC into my regulator? Lenny


** Nooooooh !!

Get a ****ing brain.



.... Phil

"klem kedidelhopper" wrote in message
...
On Feb 20, 6:06 pm, amdx wrote:
On 2/19/2013 5:01 PM, klem kedidelhopper wrote:







On Feb 19, 12:51 am, "Phil Allison" wrote:


With a full-wave bridge rectifier, if you use an 8300 uF capacitor,
the
volts of ripple will equal the amps of load current. (Don Lancaster
taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a
full
wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1
amp.

.... Phil

This is some great advice. Thank you everyone for all your input. I
especially like the diode and Zener ideas. They/re cheap and should
work well. I've got lots of diodes around here and I'm going to
experiment with them. However after all this I remembered that the
transformer is center tapped! At the time, I simply wire nutted the
tap, tucked it down under the transformer and forgot about it. How
this simple fact and the possibilities it presents slipped my mind is
anyone's guess. I've been building this thing in the evenings after my
usual work and perhaps I've been tired. The transformer measures 25.2V
at 120V line unloaded. Loading the transformer in this fashion will
probably unbalance it somewhat and drop the 24V a bit as well but I
don't see it as a real problem. I should be able to use one side of
the secondary and the tap now get at least 15 -17 VDC out of the
bridge and filter, and that will provide a healthier input to my
regulator. Lenny

No need to unbalance the transformer, just use 2 diodes and use the
center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:
"klem kedidelhopper" wrote in message

...

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:















On Feb 19, 12:51 am, "Phil Allison" wrote:





With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. (Don Lancaster

taught me this.)



** The correct value is 6300uF.



Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.



For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.



.... Phil



This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny



No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,



http://metroamp.com/wiki/index.php/F..._Tap_Rectifier



Mikek



But won't that still give me 36VDC into my regulator? Lenny



No, it won't



Arfa



I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

I looked over what I had previously written and perhaps I didn’t explain
this properly. I have a 24VCT transformer. The transformer needs to power
24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried
using a bridge directly off the 24V winding. As soon as I connected up the
filter cap the DC output went to 35V, which is probably too high to feed my
12V regulator.

Several people came up with some great ideas to address this, and I then
realized that I had a center tap that was not being used. So I connected my
bridge across the center tap and one side of the secondary. This time the
13VAC when FW rectified using the bridge went to about 19VDC, which is a
safe input to the regulator. Although this worked, I wasn’t happy about
unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the
full secondary output with two diodes, cathodes tied together, (typical FW
rectifier), and use the center tap as my negative return. I didn't think
that would make any difference in the output voltage from using a bridge
without the center tap however I tried it anyway. As I suspected it would,
the DC output again went to 35V when I connected the filter up.


** Center tap transformers are ALWAYS double the AC voltage of bridge
rectifier ones for the same final DC voltage.

You have made a dopey wiring error.


..... Phil

I looked over what I had previously written and perhaps I didn’t explain
this properly. I have a 24VCT transformer. The transformer needs to power
24VAC equipment as well as 12VDC equipment. For the DC circuit I first
tried using a bridge directly off the 24V winding. As soon as I connected
up the filter cap the DC output went to 35V, which is probably too high to
feed my 12V regulator.

Several people came up with some great ideas to address this, and I then
realized that I had a center tap that was not being used. So I connected
my bridge across the center tap and one side of the secondary. This time
the 13VAC when FW rectified using the bridge went to about 19VDC, which is
a safe input to the regulator. Although this worked, I wasn’t happy about
unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the
full secondary output with two diodes, cathodes tied together, (typical FW
rectifier), and use the center tap as my negative return. I didn't think
that would make any difference in the output voltage from using a bridge
without the center tap however I tried it anyway. As I suspected it would,
the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is
catching up with me. I guess I just don't understand the explanation of
how to get a lower voltage to my DC regulator by what I thought sounded
like just configuring the transformer properly. So at the risk of sounding
like a complete idiot here can someone please explain this further?
Thanks, Lenny



OK. A bit clearer now. What you were originally proposing now seems OK
provided that the 24v AC circuit is completely floating with respect to
anything that you want to do at 12v - ie driving your regulator input - and
the current that your 12v circuit is going to draw, is not a lot compared to
the rating of the transformer, and the current that the AC circuit is going
to draw is not near to the transformer's rating, then you should get away
with running in the proposed 'unbalanced' way. I have done such things over
the years, and gotten away with it. Although it's not very good practice,
transformers are pretty forgiving components. I guess that the simple answer
is try it, and see what happens ?

Arfa

On Feb 26, 5:16*am, "Arfa Daily" wrote:
I looked over what I had previously written and perhaps I didn’t explain
this properly. I have a 24VCT transformer. The transformer needs to power
24VAC equipment as well as 12VDC equipment. For the DC circuit I first
tried using a bridge directly off the 24V winding. As soon as I connected
up the filter cap the DC output went to 35V, which is probably too high to
feed my 12V regulator.

Several people came up with some great ideas to address this, and I then
realized that I had a center tap that was not being used. So I connected
my bridge across the center tap and one side of the secondary. This time
the 13VAC when FW rectified using the bridge went to about 19VDC, which is
a safe input to the regulator. Although this worked, I wasn’t happy about
unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the
full secondary output with two diodes, cathodes tied together, (typical FW
rectifier), and use the center tap as my negative return. I didn't think
that would make any difference in the output voltage from using a bridge
without the center tap however I tried it anyway. As I suspected it would,
the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is
catching up with me. I guess I just don't understand the explanation of
how to get a lower voltage to my DC regulator by what I thought sounded
like just configuring the transformer properly. So at the risk of sounding
like a complete idiot here can someone please explain this further?
Thanks, Lenny

OK. A bit clearer now. What you were originally proposing now seems OK
provided that the 24v AC circuit is completely floating with respect to
anything that you want to do at 12v - ie driving your regulator input - and
the current that your 12v circuit is going to draw, is not a lot compared to
the rating of the transformer, and the current that the AC circuit is going
to draw is not near to the transformer's rating, then you should get away
with running in the proposed 'unbalanced' way. I have done such things over
the years, and gotten away with it. Although it's not very good practice,
transformers are pretty forgiving components. I guess that the simple answer
is try it, and see what happens ?

Arfa

I'm grateful for everyone's input, and although this power supply as
well as the equipment that it will power is to be used in a family
business, never the less I'm trying to build this as "technically
correct" as possible. I found some 6.8V 10W stud mount zeners in the
junk box last night. I'm thinking that they may be a bit light for
this but I might try stringing two or possibly three of them together
on a heat sink as Arfa suggested or perhaps try the series diode
trick, (from Phil), both as previously mentioned on the output of the
filter and see what happens. Even though given the load it may work
fine it would be nice as Arfa said to not have to "unbalance the
transformer. Lenny

"klem kedidelhopper" wrote in message
...
On Feb 26, 5:16 am, "Arfa Daily" wrote:
I looked over what I had previously written and perhaps I didn’t
explain
this properly. I have a 24VCT transformer. The transformer needs to
power
24VAC equipment as well as 12VDC equipment. For the DC circuit I first
tried using a bridge directly off the 24V winding. As soon as I
connected
up the filter cap the DC output went to 35V, which is probably too high
to
feed my 12V regulator.

Several people came up with some great ideas to address this, and I
then
realized that I had a center tap that was not being used. So I
connected
my bridge across the center tap and one side of the secondary. This
time
the 13VAC when FW rectified using the bridge went to about 19VDC, which
is
a safe input to the regulator. Although this worked, I wasn’t happy
about
unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off
the
full secondary output with two diodes, cathodes tied together, (typical
FW
rectifier), and use the center tap as my negative return. I didn't
think
that would make any difference in the output voltage from using a
bridge
without the center tap however I tried it anyway. As I suspected it
would,
the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is
catching up with me. I guess I just don't understand the explanation of
how to get a lower voltage to my DC regulator by what I thought sounded
like just configuring the transformer properly. So at the risk of
sounding
like a complete idiot here can someone please explain this further?
Thanks, Lenny

OK. A bit clearer now. What you were originally proposing now seems OK
provided that the 24v AC circuit is completely floating with respect to
anything that you want to do at 12v - ie driving your regulator input -
and
the current that your 12v circuit is going to draw, is not a lot compared
to
the rating of the transformer, and the current that the AC circuit is
going
to draw is not near to the transformer's rating, then you should get away
with running in the proposed 'unbalanced' way. I have done such things
over
the years, and gotten away with it. Although it's not very good practice,
transformers are pretty forgiving components. I guess that the simple
answer
is try it, and see what happens ?

Arfa

I'm grateful for everyone's input, and although this power supply as
well as the equipment that it will power is to be used in a family
business, never the less I'm trying to build this as "technically
correct" as possible. I found some 6.8V 10W stud mount zeners in the
junk box last night. I'm thinking that they may be a bit light for
this but I might try stringing two or possibly three of them together
on a heat sink as Arfa suggested or perhaps try the series diode
trick, (from Phil), both as previously mentioned on the output of the
filter and see what happens. Even though given the load it may work
fine it would be nice as Arfa said to not have to "unbalance the
transformer. Lenny

That would make the zeners good for about an amp and a half tops. Bear in
mind that in order to string them in series on a heatsink, the studs will
need to be insulated from it. These stud mount diodes often come with
insulating mounting hardware. I think as an initial try-out, I'd just put
one in the line and see what you get. Should knock you down to something
just above 28 v from your 35 v starting point. Let us know how you get on

Arfa

wrote:

On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

...


On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. (Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... Phil

This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

Mikek

But won't that still give me 36VDC into my regulator? Lenny



No, it won't



Arfa



I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

Your problem is simple

If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

The center tap would be your common for the 12 volt supply, one diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..




o o
| |
| | to your 24V AC device
| |
| |
| |
| |
| |
| |
+ | |
--------------+. ,-o---------|-----|-+--+
)|( | | 18DC
)|( | +------++
Line Voltage -. ,-+----+| | | ===
)|( === | | /-\
)|( GND | | |
+-------------+-' '+----------o-----|+----+ ===
GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


Unless I missed something, I don't see a problem?
Jamie

On Feb 26, 6:39*pm, Jamie
t wrote:
wrote:
On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

...

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. *(Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... *Phil

This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! *At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

* No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

* * * * * * * * * * * * * * * *Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is *a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

* Your problem is simple

* *If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

* *The center tap would be your common for the 12 volt supply, one diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

* As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

* * * * * * * * * * * * * * * * * o * * * *o
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * | to your 24V AC device
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * + * * * * * * * * * | * * * * |
* * * * * * * *--------------+. ,-o---------|-----|-+--+
* * * * * * * * * * * * * * * )|( * * * * * | * * * * * | * *18DC
* * * * * * * * * * * * * * * )|( * * * * * | * * * * * +------++
* * * * *Line Voltage * * * *-. ,-+----+| * | * * * * * | * * ===
* * * * * * * * * * * * * * * )|( * * *=== *| * * * * * | * * /-\
* * * * * * * * * * * * * * * )|( * * *GND *| * * * * * | * * *|
* * * * * * * +-------------+-' '+----------o-----|+----+ * *===
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

* Unless I missed something, I don't see a problem?
Jamie

On Feb 26, 6:39*pm, Jamie
t wrote:
wrote:
On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

...

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. *(Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... *Phil

This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! *At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

* No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

* * * * * * * * * * * * * * * *Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is *a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

* Your problem is simple

* *If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

* *The center tap would be your common for the 12 volt supply, one diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

* As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

* * * * * * * * * * * * * * * * * o * * * *o
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * | to your 24V AC device
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * + * * * * * * * * * | * * * * |
* * * * * * * *--------------+. ,-o---------|-----|-+--+
* * * * * * * * * * * * * * * )|( * * * * * | * * * * * | * *18DC
* * * * * * * * * * * * * * * )|( * * * * * | * * * * * +------++
* * * * *Line Voltage * * * *-. ,-+----+| * | * * * * * | * * ===
* * * * * * * * * * * * * * * )|( * * *=== *| * * * * * | * * /-\
* * * * * * * * * * * * * * * )|( * * *GND *| * * * * * | * * *|
* * * * * * * +-------------+-' '+----------o-----|+----+ * *===
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

* Unless I missed something, I don't see a problem?
Jamie

I know it's difficult when we're only using text but I just can't
understand, (read that is) your schematic. From your description
though it sounds like it might be the FW arrangement I've already
tried and got 35V out of. Is there any way to make that appear any
more readable? Lenny

On Feb 26, 6:39*pm, Jamie
t wrote:
wrote:
On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

...

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. *(Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... *Phil

This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! *At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

* No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

* * * * * * * * * * * * * * * *Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is *a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

* Your problem is simple

* *If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

* *The center tap would be your common for the 12 volt supply, one diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

* As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

* * * * * * * * * * * * * * * * * o * * * *o
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * | to your 24V AC device
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * * | * * * * |
* * * * * * * + * * * * * * * * * | * * * * |
* * * * * * * *--------------+. ,-o---------|-----|-+--+
* * * * * * * * * * * * * * * )|( * * * * * | * * * * * | * *18DC
* * * * * * * * * * * * * * * )|( * * * * * | * * * * * +------++
* * * * *Line Voltage * * * *-. ,-+----+| * | * * * * * | * * ===
* * * * * * * * * * * * * * * )|( * * *=== *| * * * * * | * * /-\
* * * * * * * * * * * * * * * )|( * * *GND *| * * * * * | * * *|
* * * * * * * +-------------+-' '+----------o-----|+----+ * *===
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

* Unless I missed something, I don't see a problem?
Jamie

I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny

klem kedidelhopper wrote:

On Feb 26, 6:39 pm, Jamie
t wrote:
wrote:
On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

...

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. (Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... Phil

This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

Your problem is simple

If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

The center tap would be your common for the 12 volt supply, one diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

o o
| |
| | to your 24V AC device
| |
| |
| |
| |
| |
| |
+ | |
--------------+. ,-o---------|-----|-+--+
)|( | | 18DC
)|( | +------++
Line Voltage -. ,-+----+| | | ===
)|( === | | /-\
)|( GND | | |
+-------------+-' '+----------o-----|+----+ ===
GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

Unless I missed something, I don't see a problem?
Jamie

I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny

Copy it and paste it into Wordpad, with the Courier New font.
Google Groups uses a proportional font which screws up drawings.

klem kedidelhopper wrote:

On Feb 26, 6:39 pm, Jamie
t wrote:

wrote:

On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

...

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. (Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... Phil

This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

Your problem is simple

If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

The center tap would be your common for the 12 volt supply, one diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

o o
| |
| | to your 24V AC device
| |
| |
| |
| |
| |
| |
+ | |
--------------+. ,-o---------|-----|-+--+
)|( | | 18DC
)|( | +------++
Line Voltage -. ,-+----+| | | ===
)|( === | | /-\
)|( GND | | |
+-------------+-' '+----------o-----|+----+ ===
GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

Unless I missed something, I don't see a problem?
Jamie


I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny

THe FW you tried was a 4 diode bridge, this is a 2 diode full wave config.
THe difference being is, the CT (Center tap) is your common, the
negative terminal and where the Cathode of the 2 diodes come together
are the (+) terminal which will be around 18VDC with Cap attached

If you truly do have a 24 volt transformer with CT, this means you'll
get 12 Volts from each outer leg with respect to the CT. Because they
are out of phase with each other (180), this is a split phase
configuration, like seen in residential pole pigs in the USA.

If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT

Jamie

On Feb 27, 6:21*pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 26, 6:39 pm, Jamie
t wrote:

wrote:

On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

....

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. *(Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... *Phil

This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! *At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

*No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

* * * * * * * * * * * * * * * Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is *a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

*Your problem is simple

* If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

* The center tap would be your common for the 12 volt supply, one diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

*As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

* * * * * * * * * * * * * * * * *o * * * *o
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * | to your 24V AC device
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * *+ * * * * * * * * * | * * * * |
* * * * * * * --------------+. ,-o---------|-----|-+--+
* * * * * * * * * * * * * * *)|( * * * * * | * * * * * | * *18DC
* * * * * * * * * * * * * * *)|( * * * * * | * * * * * +------++
* * * * Line Voltage * * * *-. ,-+----+| * | * * * * * | * * ===
* * * * * * * * * * * * * * *)|( * * *=== *| * * * * * | * * /-\
* * * * * * * * * * * * * * *)|( * * *GND *| * * * * * | * * *|
* * * * * * *+-------------+-' '+----------o-----|+----+ * *===
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

*Unless I missed something, I don't see a problem?
Jamie

I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny

* THe FW you tried was a 4 diode bridge, this is a 2 diode full wave config.
* *THe difference being is, the CT (Center tap) is your common, the
negative terminal and where the Cathode of the 2 diodes come together
are the (+) terminal which will be around 18VDC with Cap attached

* If you truly do have a 24 volt transformer with CT, this means you'll
get 12 Volts from each outer leg with respect to the CT. Because they
are out of phase with each other (180), this is a split phase
configuration, like seen in residential pole pigs in the USA.

* If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT

Jamie

What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT as common connected to
an output filter will yield 18VDC. A 24 V transformer under the same
circumstances will produce 36VDC. I've tried it and it does. Lenny.

"klem kedidelhopper"

If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT

Jamie

What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT as common connected to
an output filter will yield 18VDC.


** ****ing BULL **** !!!!!!!!!!!!!!


I've tried it and it does


** Then explain how every other person on the planet get a different result
??

Using a CT as common halves the voltage from the tranny and so halves the DC
voltage as well.

My god you are ****ing stupid.



..... Phil

Jamie

What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT as common connected to
an output filter will yield 18VDC. A 24 V transformer under the same
circumstances will produce 36VDC. I've tried it and it does. Lenny.

That's not true, I'm afraid. Using the CT as common, and having a diode on
each 'hot' limb of the winding will produce half the voltage of hanging a
bridge across the same two limbs, and *not* using the centre tap. If you do
the same, but continue to use the CT as 'common', then you will get +18v at
the "+" terminal of the bridge, and -18v at the "-" terminal ...

Arfa

klem kedidelhopper wrote:

On Feb 27, 6:21 pm, Jamie
t wrote:

klem kedidelhopper wrote:

On Feb 26, 6:39 pm, Jamie
et wrote:

wrote:

On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

...

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. (Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... Phil

This is some great advice. Thank you everyone for all your input. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

Your problem is simple

If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

The center tap would be your common for the 12 volt supply, one diode

from each outter leg joined together to form a full wave. The diode

output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

o o
| |
| | to your 24V AC device
| |
| |
| |
| |
| |
| |
+ | |
--------------+. ,-o---------|-----|-+--+
)|( | | 18DC
)|( | +------++
Line Voltage -. ,-+----+| | | ===
)|( === | | /-\
)|( GND | | |
+-------------+-' '+----------o-----|+----+ ===
GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

Unless I missed something, I don't see a problem?
Jamie

I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny

THe FW you tried was a 4 diode bridge, this is a 2 diode full wave config.
THe difference being is, the CT (Center tap) is your common, the
negative terminal and where the Cathode of the 2 diodes come together
are the (+) terminal which will be around 18VDC with Cap attached

If you truly do have a 24 volt transformer with CT, this means you'll
get 12 Volts from each outer leg with respect to the CT. Because they
are out of phase with each other (180), this is a split phase
configuration, like seen in residential pole pigs in the USA.

If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT

Jamie


What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT as common connected to
an output filter will yield 18VDC. A 24 V transformer under the same
circumstances will produce 36VDC. I've tried it and it does. Lenny.

absolutely not, you are greatly mistaken or you don't know what you have
for a transformer.

THe transformer is spec'ed out via its outer legs, the highest
voltage. The CT is the voltage divider. when you use the CT as your
common, the supply will yield only half of the rated voltage of that
transformer and is when you use only 2 diodes from the outer legs to
form a full wave into a single node where this becomes the (+) terminal.


With a real 24V Transformer with CT and using the CT as your common
(-) terminal, you will get 12*1.414 = 16.968-0.650 = 16.318VDC at the
peak of a marginal ripple.


You have your logic mixed up

Jamie

On Feb 28, 7:21*pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 27, 6:21 pm, Jamie
t wrote:

klem kedidelhopper wrote:

On Feb 26, 6:39 pm, Jamie
et wrote:

wrote:

On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:

"klem kedidelhopper" wrote in message

...

On Feb 20, 6:06 pm, amdx wrote:

On 2/19/2013 5:01 PM, klem kedidelhopper wrote:

On Feb 19, 12:51 am, "Phil Allison" wrote:



With a full-wave bridge rectifier, if you use an 8300 uF capacitor,

the

volts of ripple will equal the amps of load current. *(Don Lancaster

taught me this.)

** The correct value is 6300uF.

Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a

full

wave rectifier at 60Hz.

For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1

amp.

.... *Phil

This is some great advice. Thank you everyone for all your input.. I

especially like the diode and Zener ideas. They/re cheap and should

work well. I've got lots of diodes around here and I'm going to

experiment with them. However after all this I remembered that the

transformer is center tapped! *At the time, I simply wire nutted the

tap, tucked it down under the transformer and forgot about it. How

this simple fact and the possibilities it presents slipped my mind is

anyone's guess. I've been building this thing in the evenings after my

usual work and perhaps I've been tired. The transformer measures 25.2V

at 120V line unloaded. Loading the transformer in this fashion will

probably unbalance it somewhat and drop the 24V a bit as well but I

don't see it as a real problem. I should be able to use one side of

the secondary and the tap now get at least 15 -17 VDC out of the

bridge and filter, and that will provide a healthier input to my

regulator. Lenny

No need to unbalance the transformer, just use 2 diodes and use the

center tap as negative. See Here,

http://metroamp.com/wiki/index.php/F..._Tap_Rectifier

* * * * * * * * * * * * * * *Mikek

But won't that still give me 36VDC into my regulator? Lenny

No, it won't

Arfa

I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is *a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny

Your problem is simple

*If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.

*The center tap would be your common for the 12 volt supply, one diode

from each outter leg joined together to form a full wave. The diode

output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.

As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

* * * * * * * * * * * * * * * * o * * * *o
* * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * | * * * * | to your 24V AC device
* * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * | * * * * |
* * * * * * * * * * * * * * * * | * * * * |
* * * * * * + * * * * * * * * * | * * * * |
* * * * * * *--------------+. ,-o---------|-----|-+--+
* * * * * * * * * * * * * * )|( * * * * * | * * * * * | * *18DC
* * * * * * * * * * * * * * )|( * * * * * | * * * * * +------++
* * * *Line Voltage * * * *-. ,-+----+| * | * * * * * | * * ===
* * * * * * * * * * * * * * )|( * * *=== *| * * * * * | * * /-\
* * * * * * * * * * * * * * )|( * * *GND *| * * * * * | * * *|
* * * * * * +-------------+-' '+----------o-----|+----+ * *===
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

Unless I missed something, I don't see a problem?
Jamie

I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny

*THe FW you tried was a 4 diode bridge, this is a 2 diode full wave config.
* THe difference being is, the CT (Center tap) is your common, the
negative terminal and where the Cathode of the 2 diodes come together
are the (+) terminal which will be around 18VDC with Cap attached

*If you truly do have a 24 volt transformer with CT, this means you'll
get 12 Volts from each outer leg with respect to the CT. Because they
are out of phase with each other (180), this is a split phase
configuration, like seen in residential pole pigs in the USA.

*If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT

Jamie

What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT *as common connected to
an output filter will yield 18VDC. A 24 V transformer under the same
circumstances will produce 36VDC. I've tried it and it does. Lenny.

absolutely not, you are greatly mistaken or you don't know what you have
for a transformer.

* *THe transformer is spec'ed out via its outer legs, the highest
voltage. The CT is the voltage divider. when you use the CT as your
common, the supply will yield only half of the rated voltage of that
transformer and is when you use only 2 diodes from the outer legs to
form a full wave into a single node where this becomes the (+) terminal.

* *With a real 24V Transformer with CT and using the CT as your common
(-) terminal, you will get 12*1.414 = 16.968-0.650 = 16.318VDC at the
peak of a marginal ripple.

* *You have your logic mixed up

Jamie

What I decided to do is use the full secondary and employ the two 6.8V
zeners in series with the cap. That arrangement gives me 21V out of my
power supply to feed the 12V regulator. As far as I can tell the other
methods would be using the center tap and one side of the transformer,
a method which I didn't want to use. I tried this full secondary
arrangement with a 1.2A load and the 21V across the zeners holds
pretty steady. I have a nice heat sink for the two diodes, (they will
be insulated) that will be going into the box as well. Thanks everyone
who advised and helped me on this project. Lenny

"klem kedidelhopper"

What I decided to do is use the full secondary and employ the two 6.8V
zeners in series with the cap. That arrangement gives me 21V out of my
power supply to feed the 12V regulator.


** He sure is one stubborn SOB - ain't he?




..... Phil

"Phil Allison" wrote in message
...

"klem kedidelhopper"

What I decided to do is use the full secondary and employ the two 6.8V
zeners in series with the cap. That arrangement gives me 21V out of my
power supply to feed the 12V regulator.


** He sure is one stubborn SOB - ain't he?




.... Phil





Well, as long as it works ok for him, and he understands it, I guess it's a
result in the end ...

Arfa

"Arfa Daily"
"Phil Allison"
"klem kedidelhopper"

What I decided to do is use the full secondary and employ the two 6.8V
zeners in series with the cap. That arrangement gives me 21V out of my
power supply to feed the 12V regulator.


** He sure is one stubborn SOB - ain't he?


Well, as long as it works ok for him, and he understands it, I guess it's
a result in the end ...


** All the fool had to do was LOOK at his wiring and spot the mistake.

The CT wire was swapped with one of the ends.

FFS, what a annoying moron.

Phallus'n wrote:



** He sure is one stubborn SOB - ain't he?


.... Phil

Snort.... No one is more stubborn than Phallus'n. No one. Phil stubbornly hangs on to an argument when he's been long proven wrong. His stubbornness is so well known in fact that he was given a guest cameo on a sitcom. He's the one on the right. Enjoy:

http://www.youtube.com/watch?v=fqs9DYisSsg

klem kedidelhopper wrote:
On Feb 28, 7:21 pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 27, 6:21 pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 26, 6:39 pm, Jamie
t wrote:
wrote:
On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:
"klem kedidelhopper" wrote in message
...
On Feb 20, 6:06 pm, amdx wrote:
On 2/19/2013 5:01 PM, klem kedidelhopper wrote:
On Feb 19, 12:51 am, "Phil Allison" wrote:

With a full-wave bridge rectifier, if you use an 8300 uF capacitor,
the
volts of ripple will equal the amps of load current. (Don Lancaster
taught me this.)
** The correct value is 6300uF.
Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a
full
wave rectifier at 60Hz.
For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1
amp.
.... Phil
This is some great advice. Thank you everyone for all your input.. I
especially like the diode and Zener ideas. They/re cheap and should
work well. I've got lots of diodes around here and I'm going to
experiment with them. However after all this I remembered that the
transformer is center tapped! At the time, I simply wire nutted the
tap, tucked it down under the transformer and forgot about it. How
this simple fact and the possibilities it presents slipped my mind is
anyone's guess. I've been building this thing in the evenings after my
usual work and perhaps I've been tired. The transformer measures 25.2V
at 120V line unloaded. Loading the transformer in this fashion will
probably unbalance it somewhat and drop the 24V a bit as well but I
don't see it as a real problem. I should be able to use one side of
the secondary and the tap now get at least 15 -17 VDC out of the
bridge and filter, and that will provide a healthier input to my
regulator. Lenny
No need to unbalance the transformer, just use 2 diodes and use the
center tap as negative. See Here,
http://metroamp.com/wiki/index.php/F..._Tap_Rectifier
Mikek
But won't that still give me 36VDC into my regulator? Lenny
No, it won't
Arfa
I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried using a bridge directly off the 24V winding. As soon as I connected up the filter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.
Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.
Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.
I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny
Your problem is simple
If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave config.
The center tap would be your common for the 12 volt supply, one diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
could do that. Use a LDO type.
As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..
o o
| |
| | to your 24V AC device
| |
| |
| |
| |
| |
| |
+ | |
--------------+. ,-o---------|-----|-+--+
)|( | | 18DC
)|( | +------++
Line Voltage -. ,-+----+| | | ===
)|( === | | /-\
)|( GND | | |
+-------------+-' '+----------o-----|+----+ ===
GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)
Unless I missed something, I don't see a problem?
Jamie
I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny
THe FW you tried was a 4 diode bridge, this is a 2 diode full wave config.
THe difference being is, the CT (Center tap) is your common, the
negative terminal and where the Cathode of the 2 diodes come together
are the (+) terminal which will be around 18VDC with Cap attached
If you truly do have a 24 volt transformer with CT, this means you'll
get 12 Volts from each outer leg with respect to the CT. Because they
are out of phase with each other (180), this is a split phase
configuration, like seen in residential pole pigs in the USA.
If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT
Jamie
What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT as common connected to
an output filter will yield 18VDC. A 24 V transformer under the same
circumstances will produce 36VDC. I've tried it and it does. Lenny.
absolutely not, you are greatly mistaken or you don't know what you have
for a transformer.

THe transformer is spec'ed out via its outer legs, the highest
voltage. The CT is the voltage divider. when you use the CT as your
common, the supply will yield only half of the rated voltage of that
transformer and is when you use only 2 diodes from the outer legs to
form a full wave into a single node where this becomes the (+) terminal.

With a real 24V Transformer with CT and using the CT as your common
(-) terminal, you will get 12*1.414 = 16.968-0.650 = 16.318VDC at the
peak of a marginal ripple.

You have your logic mixed up

Jamie

What I decided to do is use the full secondary and employ the two 6.8V
zeners in series with the cap. That arrangement gives me 21V out of my
power supply to feed the 12V regulator. As far as I can tell the other
methods would be using the center tap and one side of the transformer,
a method which I didn't want to use. I tried this full secondary
arrangement with a 1.2A load and the 21V across the zeners holds
pretty steady. I have a nice heat sink for the two diodes, (they will
be insulated) that will be going into the box as well. Thanks everyone
who advised and helped me on this project. Lenny

Until one of your zener diodes short out (a zener will eventually short
out due to inrush current issues is my guess) and you then punch around
19V to the circuit that wants 12....not a good design as a result, more
of a hack.

John :-#(#

"John Robertson" wrote in message
...

klem kedidelhopper wrote:
On Feb 28, 7:21 pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 27, 6:21 pm, Jamie
t wrote:
klem kedidelhopper wrote:
On Feb 26, 6:39 pm, Jamie
t wrote:
wrote:
On Sunday, February 24, 2013 10:17:29 AM UTC-5, Arfa Daily wrote:
"klem kedidelhopper" wrote in
message
...
On Feb 20, 6:06 pm, amdx wrote:
On 2/19/2013 5:01 PM, klem kedidelhopper wrote:
On Feb 19, 12:51 am, "Phil Allison" wrote:

With a full-wave bridge rectifier, if you use an 8300 uF
capacitor,
the
volts of ripple will equal the amps of load current. (Don
Lancaster
taught me this.)
** The correct value is 6300uF.
Such a cap will drop 1 volt in 6.3mS, the typical discharge
time in a
full
wave rectifier at 60Hz.
For 50Hz full wave supplies, the value is 7500uF for 1V ripple
at 1
amp.
.... Phil
This is some great advice. Thank you everyone for all your
input.. I
especially like the diode and Zener ideas. They/re cheap and
should
work well. I've got lots of diodes around here and I'm going to
experiment with them. However after all this I remembered that
the
transformer is center tapped! At the time, I simply wire nutted
the
tap, tucked it down under the transformer and forgot about it.
How
this simple fact and the possibilities it presents slipped my
mind is
anyone's guess. I've been building this thing in the evenings
after my
usual work and perhaps I've been tired. The transformer measures
25.2V
at 120V line unloaded. Loading the transformer in this fashion
will
probably unbalance it somewhat and drop the 24V a bit as well
but I
don't see it as a real problem. I should be able to use one side
of
the secondary and the tap now get at least 15 -17 VDC out of the
bridge and filter, and that will provide a healthier input to my
regulator. Lenny
No need to unbalance the transformer, just use 2 diodes and use
the
center tap as negative. See Here,
http://metroamp.com/wiki/index.php/F..._Tap_Rectifier
Mikek
But won't that still give me 36VDC into my regulator? Lenny
No, it won't
Arfa
I looked over what I had previously written and perhaps I didn’t
explain this properly. I have a 24VCT transformer. The transformer
needs to power 24VAC equipment as well as 12VDC equipment. For the
DC circuit I first tried using a bridge directly off the 24V
winding. As soon as I connected up the filter cap the DC output went
to 35V, which is probably too high to feed my 12V regulator.
Several people came up with some great ideas to address this, and I
then realized that I had a center tap that was not being used. So I
connected my bridge across the center tap and one side of the
secondary. This time the 13VAC when FW rectified using the bridge
went to about 19VDC, which is a safe input to the regulator.
Although this worked, I wasn’t happy about unbalancing the
transformer this way so I posted my results.
Unless I seriously misunderstood It was suggested here that I come
off the full secondary output with two diodes, cathodes tied
together, (typical FW rectifier), and use the center tap as my
negative return. I didn't think that would make any difference in
the output voltage from using a bridge without the center tap
however I tried it anyway. As I suspected it would, the DC output
again went to 35V when I connected the filter up.
I've been playing with power supplies all my life and maybe my age
is catching up with me. I guess I just don't understand the
explanation of how to get a lower voltage to my DC regulator by what
I thought sounded like just configuring the transformer properly. So
at the risk of sounding like a complete idiot here can someone
please explain this further? Thanks, Lenny
Your problem is simple
If the 24V AC devices does not come in contact with the common or
ground of this supply you have, you can use a 2 diode full wave
config.
The center tap would be your common for the 12 volt supply, one
diode
from each outter leg joined together to form a full wave. The diode
output alogn with using the center tap as your common will give you
12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess
you
could do that. Use a LDO type.
As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd
or
DC out..
o o
| |
| | to your 24V AC device
| |
| |
| |
| |
| |
| |
+ | |
--------------+. ,-o---------|-----|-+--+
)|( | | 18DC
)|( | +------++
Line Voltage -. ,-+----+| | | ===
)|( === | | /-\
)|( GND | | |
+-------------+-' '+----------o-----|+----+ ===
GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)
Unless I missed something, I don't see a problem?
Jamie
I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny
THe FW you tried was a 4 diode bridge, this is a 2 diode full wave
config.
THe difference being is, the CT (Center tap) is your common, the
negative terminal and where the Cathode of the 2 diodes come together
are the (+) terminal which will be around 18VDC with Cap attached
If you truly do have a 24 volt transformer with CT, this means you'll
get 12 Volts from each outer leg with respect to the CT. Because they
are out of phase with each other (180), this is a split phase
configuration, like seen in residential pole pigs in the USA.
If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT
Jamie
What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT as common connected to
an output filter will yield 18VDC. A 24 V transformer under the same
circumstances will produce 36VDC. I've tried it and it does. Lenny.
absolutely not, you are greatly mistaken or you don't know what you have
for a transformer.

THe transformer is spec'ed out via its outer legs, the highest
voltage. The CT is the voltage divider. when you use the CT as your
common, the supply will yield only half of the rated voltage of that
transformer and is when you use only 2 diodes from the outer legs to
form a full wave into a single node where this becomes the (+) terminal.

With a real 24V Transformer with CT and using the CT as your common
(-) terminal, you will get 12*1.414 = 16.968-0.650 = 16.318VDC at the
peak of a marginal ripple.

You have your logic mixed up

Jamie

What I decided to do is use the full secondary and employ the two 6.8V
zeners in series with the cap. That arrangement gives me 21V out of my
power supply to feed the 12V regulator. As far as I can tell the other
methods would be using the center tap and one side of the transformer,
a method which I didn't want to use. I tried this full secondary
arrangement with a 1.2A load and the 21V across the zeners holds
pretty steady. I have a nice heat sink for the two diodes, (they will
be insulated) that will be going into the box as well. Thanks everyone
who advised and helped me on this project. Lenny

Until one of your zener diodes short out (a zener will eventually short
out due to inrush current issues is my guess) and you then punch around
19V to the circuit that wants 12....not a good design as a result, more
of a hack.

John :-#(#



Well then you want another protection Zener shunted on the 12v line.



Gareth.

On Wed, 27 Feb 2013 21:24:41 -0800 (PST), klem kedidelhopper
wrote:


*As for the 24 v AC, the outter legs will supply that to the out board
device. Make sure the AC legs do not come in contact with either grd or
DC out..

* * * * * * * * * * * * * * * * *o * * * *o
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * | to your 24V AC device
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * * * * * * * * * * * *| * * * * |
* * * * * * *+ * * * * * * * * * | * * * * |
* * * * * * * --------------+. ,-o---------|-----|-+--+
* * * * * * * * * * * * * * *)|( * * * * * | * * * * * | * *18DC
* * * * * * * * * * * * * * *)|( * * * * * | * * * * * +------++
* * * * Line Voltage * * * *-. ,-+----+| * | * * * * * | * * ===
* * * * * * * * * * * * * * *)|( * * *=== *| * * * * * | * * /-\
* * * * * * * * * * * * * * *)|( * * *GND *| * * * * * | * * *|
* * * * * * *+-------------+-' '+----------o-----|+----+ * *===
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *GND
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

*Unless I missed something, I don't see a problem?
Jamie

I'm not sure if my last reply worked so I'll try it again. I know that
it's very difficult when all we can use here is text but I'm afraid
that I just can't understand, (read) your schematic. From you
description however it appears to be the FW arrangement that I've
already tried and got 35 V out of after I connected up the filter. Is
there any way to make your diagram any more readable in this medium?
Thanks, Lenny

* THe FW you tried was a 4 diode bridge, this is a 2 diode full wave config.
* *THe difference being is, the CT (Center tap) is your common, the
negative terminal and where the Cathode of the 2 diodes come together
are the (+) terminal which will be around 18VDC with Cap attached

* If you truly do have a 24 volt transformer with CT, this means you'll
get 12 Volts from each outer leg with respect to the CT. Because they
are out of phase with each other (180), this is a split phase
configuration, like seen in residential pole pigs in the USA.

* If you have done this and gotten 36 volts, then you actually have a
48V transformer with CT

Jamie

What you're referring to is a 12VCT transformer. Connecting either a
bridge and not using a center tap or just two diodes across the full
secondary with cathodes tied together using CT as common connected to
an output filter will yield 18VDC. A 24 V transformer under the same
circumstances will produce 36VDC. I've tried it and it does. Lenny.


Two questions:

Do you have LTSpice?

Can you read alt.binaries.schematics.electronic?

If the answer in no in both cases, get LTSpice.

The .asc files are very handy for this kind of discussion.

?-)

Until one of your zener diodes short out (a zener will eventually short
out due to inrush current issues is my guess) and you then punch around
19V to the circuit that wants 12....not a good design as a result, more
of a hack.

John :-#(#



Yebbut, the zeners are not dropping the supply to 12v. They are *reducing*
the input supply to a full blown 12 v regulator sub-circuit that the OP
already has. Further, that regulator is expecting smooth DC in, so probably
only has a few tens of uF hung across its input, for decoupling purposes, so
there will not be any serious inrush current through the zeners. Given that
fact, there is no reason why the zeners should ever fail, particularly as
they are high wattage stud types anyway. The technique of placing zeners in
series with voltages that you want to reduce by a steady and relatively
current-independent amount, is well known, and not uncommon in commercial
designs.

Arfa

"John-Del" wrote in message
...
Phallus'n wrote:



** He sure is one stubborn SOB - ain't he?


.... Phil

Snort.... No one is more stubborn than Phallus'n. No one. Phil
stubbornly hangs on to an argument when he's been long proven wrong. His
stubbornness is so well known in fact that he was given a guest cameo on a
sitcom. He's the one on the right. Enjoy:

http://www.youtube.com/watch?v=fqs9DYisSsg



But when he's right, he's right. And in this case, he is ...

Arfa

But when he's right, he's right. And in this case, he is ...



Arfa

That's not the point.. Phil chides the OP for being stubborn, but all one could accuse him of is being wrong.

I guess it's the internet bullying that makes Phil such a dick, and I'm sure he's a bigger tool in person. Would he act out in such a manner if he was standing directly in front of the OP? I doubt it. Over the years, I've discovered that most bullys are spineless, who hide behind a false facade of toughness (until called on it), or behind a keyboard.

If I knew that Phil acted the same way in person as he does on the internet, I might even like the guy, and buy him a beer (after pounding the snot out of him). But I'd bet almost anything that if challenged, he'd run home to mommy and hide in her basement.

"John-Del" = ****WIT TROLL

But when he's right, he's right. And in this case, he is ...


Arfa

That's not the point.. Phil chides the OP for being stubborn, but all one
could accuse him of is being wrong.


** Completely and utterly WRONG !!!

**** OFF TO HELL

- YOU RETARDED ****ING NET STALKER



.... Phil